Question

If m₁ and m₂ are roots of the equation x² + (√3 + 2) x + √3 - 1 = 0, then the area of Δ formed by the line y = m₁x, y = m₂x, y = c is

• A. $\frac{\sqrt{33} + \sqrt{11}}{4}c^2$
• B. $\frac{\sqrt{32} + \sqrt{11}}{16}c$
• C. $\frac{\sqrt{33} + \sqrt{10}}{4}c^2$
• D. $\frac{\sqrt{33} + \sqrt{21}}{4}c^3$

Answer 1

Answer: (A)

$\frac{\sqrt{33} + \sqrt{11}}{4}c^2$

Answer 2

The vertices of the triangle are $(0,0)$, $(c/m_1, c)$, and $(c/m_2, c)$. The area is $\frac{c^2}{2} \frac{|m_2 - m_1|}{|m_1 m_2|}$. From Vieta's formulas, $m_1+m_2 = -(\sqrt{3}+2)$ and $m_1m_2 = \sqrt{3}-1$. We find $(m_2-m_1)^2 = (m_1+m_2)^2 - 4m_1m_2 = 11$, so $|m_2-m_1| = \sqrt{11}$. Substituting these values and rationalizing the denominator yields $\frac{c^2(\sqrt{33}+\sqrt{11})}{4}$.